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What is the relationship between entropy and the number of possible arrangements of molecules in a system?

Short Answer

Expert verified
Entropy increases with the number of possible molecular arrangements, as shown by \( S = k \ln \Omega \).

Step by step solution

01

Understanding Entropy

Entropy is a measure of the randomness or disorder in a system. It tells us how many microscopic configurations correspond to a thermodynamic system's macroscopic state.
02

Defining Configuration

A configuration refers to a specific arrangement of molecules within a system. In thermodynamics, the macrostate of a system can have many microstates or configurations.
03

Introducing the Formula

The relationship between entropy \( S \) and the number of possible arrangements \( \Omega \) is given by Boltzmann's entropy formula: \( S = k \ln \Omega \), where \( k \) is Boltzmann's constant.
04

Interpreting the Formula

In this formula, \( \Omega \) represents the total number of possible configurations (microstates), and \( \ln \) is the natural logarithm, which helps scale the number to a usable figure for entropy.
05

Conclusion of the Relationship

The relationship shows that as the number of possible arrangements \( \Omega \) increases, the entropy \( S \) also increases, indicating greater disorder or randomness in the system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy and Disorder
Entropy provides insight into the level of disorder or randomness within a system. In thermodynamics, it's used to determine how dispersed the energy within a system is. Imagine a room with perfectly stacked boxes; this presents a low entropy or low disorder scenario.
If those boxes are randomly scattered, the disorder or entropy is higher. Thus, systems naturally progress towards a state of higher entropy or disorder over time.
Entropy is crucial in understanding however energy is distributed or transferred. It's a core concept in the second law of thermodynamics, which says that the overall entropy of an isolated system will always increase. This means that processes will naturally tend towards equilibrium and maximum entropy, reflecting maximum disorder.
In everyday language, this explains why heat flows from hot to cold and not the other way around.
Boltzmann's Entropy Formula
Boltzmann's entropy formula is a fundamental equation in statistical mechanics. It connects the microscopic properties of molecules to the macroscopic measure of entropy. The formula is expressed as:- \[ S = k \ln \Omega \]- The formula offers a quantitative approach to understanding disorder.In this equation, \( S \) represents the entropy of the system. The \( \Omega \) denotes the total number of possible microstates, or specific ways the system's molecules can be arranged.
The \( \ln \) is the natural logarithm, a mathematical function that helps translate huge numbers of microstates into a more understandable scale.The \( k \) is Boltzmann's constant, a proportionality factor in this context, providing a bridge between the microstates and entropy. With this formula, if a thermodynamic system has more possible arrangements (\( \Omega \)), it leads to a higher entropy (\( S \)), depicting more disorder.
Microstates and Macrostates
In thermodynamics, a macrostate refers to an observable condition of a system, such as temperature, pressure, or volume. In contrast, microstates are the multiple, detailed ways or configurations in which the molecules within a system can be arranged to achieve a particular macrostate.
A simple analogy would be a deck of cards. The overall ordered deck is akin to a macrostate. Every shuffle of the deck creates a new microstate, although the macrostate (the complete deck) hasn't changed. Many microstates can correspond to the same macrostate, leading to different entropy levels.
Understanding this distinction helps clarify how thermodynamic systems operate. Entropy is higher when there are many microstates possible for a given macrostate as it reflects the system's increased disorder and energy distribution.
Thermodynamic Systems
A thermodynamic system is any defined space or quantity of matter separated by boundaries where energy exchanges occur. These systems are classified as open, closed, or isolated based on their interactions with the surroundings. - **Open Systems**: Exchange both matter and energy with their surroundings. - **Closed Systems**: Exchange only energy, not matter. - **Isolated Systems**: Do not exchange energy or matter. These systems are critical in studying how energy and matter interact. For instance, in a closed system, while matter is retained, energy (such as heat) may be added or removed.
Understanding thermodynamic systems provides the basis for concepts like entropy. This framework helps to evaluate how molecules move and rearrange, contributing to the disorder and energy distribution across the system. Knowing whether a system is open, closed, or isolated affects how entropy and the number of possible microstates change over time, impacting the system’s overall behavior.

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Most popular questions from this chapter

A student looked up the \(\Delta G_{\mathrm{i}}^{\circ}, \Delta H_{\mathrm{f}}^{\circ}\), and \(\Delta S^{\circ}\) values for \(\mathrm{CO}_{2}\) in Appendix 2. Plugging these values into Equation \(18.10,\) the student found that \(\Delta G_{\mathrm{f}}^{\circ} \neq \Delta H_{\mathrm{i}}^{\circ}-\) \(T \Delta S^{\circ}\) at \(298 \mathrm{~K}\). What is wrong with this approach?

Heating copper(II) oxide at \(400^{\circ} \mathrm{C}\) does not produce any appreciable amount of Cu: $$ \mathrm{CuO}(s) \rightleftharpoons \mathrm{Cu}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta G^{\circ}=127.2 \mathrm{~kJ} / \mathrm{mol} $$ However, if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write an equation for the coupled process, and calculate the equilibrium constant for the coupled reaction.

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a lower entropy than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

From the values of \(\Delta H\) and \(\Delta S\), predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}:\) reaction A: \(\Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} ;\) reaction B: \(\Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C},\) at what temperature might it become spontaneous?

Consider two carboxylic acids (acids that contain the \(-\) COOH group \(): \mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid, \(\left.K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) and \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) (chloroacetic acid, \(K_{\mathrm{a}}=1.4 \times 10^{-3}\) ). (a) Calculate \(\Delta G^{\circ}\) for the ionization of these acids at \(25^{\circ} \mathrm{C}\). (b) From the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ},\) we see that the contributions to the \(\Delta G^{\circ}\) term are an enthalpy term \(\left(\Delta H^{\circ}\right)\) and a temperature times entropy term \(\left(T \Delta S^{\circ}\right)\). These contributions are listed here for the two acids: \begin{tabular}{lcc} & \(\Delta \boldsymbol{H}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l})\) & \(T \Delta S^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l})\) \\ \hline \(\mathrm{CH}_{3} \mathrm{COOH}\) & -0.57 & -27.6 \\\ \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) & -4.7 & -21.1 \end{tabular} Which is the dominant term in determining the value of \(\Delta G^{\circ}\) (and hence \(K_{\mathrm{a}}\) of the acid)? (c) What processes contribute to \(\Delta H^{\circ} ?\) (Consider the ionization of the acids as a Bronsted acid-base reaction.) (d) Explain why the \(T \Delta S^{\circ}\) term is more negative for \(\mathrm{CH}_{3} \mathrm{COOH} .\)

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